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3 years ago

Which of the following describes all illustrations created by freehand?

Engineering

2 answers:

ale4655 [162]3 years ago

8 0

Answer:

Sketches

Explanation:

Klio2033 [76]3 years ago

5 0

extension lines,sketches,leader lines,dimensions describes all illustrations created by freehand.

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Draw a sinusoidal signal and illustrate how quantization and sampling is handled by

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Answer: Your mums gay

Explanation: Your mums gay

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3 years ago

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Answer:

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3 years ago

Explain the two advantages and the two disadvantages of fission as an energy source.

yawa3891 [41]

Answer with Explanation:

1) The advantages of fission energy are:

a) Higher concentration of energy : Concentration of energy or the energy density is defined as the amount of energy that is produced by burning a unit mass of the fuel. The nuclear energy obtained by fission has the highest energy density among all the other natural sources of energy such as coal,gas,e.t.c.

b) Cheap source of energy : The cost at which the energy is produced by a nuclear reactor after it is operational is the lowest among all the other sources of energy such as coal, solar,e.t.c

2) The disadvantages of fission energy are:

a) Highly dangerous residue: The fuel that is left unspent is highly radioactive and thus is very dangerous. Usually the residual material is taken deep into the earth for it's disposal.

b) It has high initial costs of design and development: The cost to design a nuclear reactor and to built one after it is designed is the most among all other types of energy sources and requires highly skilled personnel for operation.

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2 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

6 0

2 years ago

I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal

tigry1 [53]

Answer:

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

where, Nₐ₀-Nₐ = is the amount in moles of A used up

Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

from conservation of mass law, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if 3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

5 0

2 years ago